Chapter 26, Problem 19P

To determine

The magnetic force on an electron.

Answer to Problem 19P

  $-(1.92\times {10}^{-13}\text{ N})\widehat{i}-(1.28\times {10}^{-13}\text{ N})\widehat{j}-(5.76\times {10}^{-13}\text{ N})\widehat{k}$

Explanation of Solution

Given:

Charge on the electron $=q=-1.6\times {10}^{-19}\text{C }$

Velocity of the particle $=\overrightarrow{v}=\left(2\times {10}^6\frac{\text{m}}{\text{s}}\right)\widehat{i}-\left(3\times {10}^6\frac{\text{m}}{\text{s}}\right)\widehat{j}$

Magnetic field $=\overrightarrow{B}=(0.80\text{ T})\widehat{i}+(0.60\text{ T})\widehat{j}-(0.40\text{ T})\widehat{k}$

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Calculation:

Magnetic force on the electron is given as

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

  $\overrightarrow{F}=(-1.6\times {10}^{-19})(((2\times {10}^6)\widehat{i}-(3\times {10}^6)\widehat{j})\times (0.80\widehat{i}+0.60\widehat{j}-0.40\widehat{k}))$

  $\overrightarrow{F}=(-1.6\times {10}^{-19})((2\times {10}^6)(0.80)(\widehat{i}\times \widehat{i})+(2\times {10}^6)(0.60)(\widehat{i}\times \widehat{j})-(2\times {10}^6)(0.40)(\widehat{i}\times \widehat{k})$

  $-(3\times {10}^6)(0.80)(\widehat{j}\times \widehat{i})-(3\times {10}^6)(0.60)(\widehat{j}\times \widehat{j})+(3\times {10}^6)(0.40)(\widehat{j}\times \widehat{k})$

  $\overrightarrow{F}=(-1.6\times {10}^{-19})((2\times {10}^6)(0.80)(0)+(2\times {10}^6)(0.60)(\widehat{k})-(2\times {10}^6)(0.40)(-\widehat{j})$

  $-(3\times {10}^6)(0.80)(-\widehat{k})-(3\times {10}^6)(0.60)(0)+(3\times {10}^6)(0.40)(\widehat{i})$

  $\overrightarrow{F}=(-1.6\times {10}^{-19})((1.2\times {10}^6)(\widehat{k})-(0.8\times {10}^6)(-\widehat{j})-(2.4\times {10}^6)(-\widehat{k})+(1.2\times {10}^6)(\widehat{i}))$

  $\overrightarrow{F}=(-1.6\times {10}^{-19})((3.6\times {10}^6)(\widehat{k})+(0.8\times {10}^6)(\widehat{j})+(1.2\times {10}^6)(\widehat{i}))$

  $\overrightarrow{F}=-(1.92\times {10}^{-13})\widehat{i}-(1.28\times {10}^{-13})\widehat{j}-(5.76\times {10}^{-13})\widehat{k}$

  $\overrightarrow{F}=-(1.92\times {10}^{-13}\text{ N})\widehat{i}-(1.28\times {10}^{-13}\text{ N})\widehat{j}-(5.76\times {10}^{-13}\text{ N})\widehat{k}$

Conclusion:

The magnetic force on the electronis $-(1.92\times {10}^{-13}\text{ N})\widehat{i}-(1.28\times {10}^{-13}\text{ N})\widehat{j}-(5.76\times {10}^{-13}\text{ N})\widehat{k}$ .